如何将 size_t 转换为 double 或 int C++

Question

My question is that

I have a size_t data, but now I want to convert it to double or int.

If I do something like

 size_t data = 99999999;
 int convertdata = data;

the compiler will report warning. because it maybe overflow.

Do you have some method like the boost or some other method to do the convert?

Answer 1

A cast, as Blaz Bratanic suggested :

size_t data = 99999999;
int convertdata = static_cast<int>(data);

is likely to silence the warning (though in principle a compiler can warn about anything it likes, even if there's a cast).

But it doesn't solve the problem that the warning was telling you about, namely that a conversion from size_t to int really could overflow.

If at all possible, design your program so you don't need to convert a size_t value to int . Just store it in a size_t variable (as you've already done) and use that.

Converting to double will not cause an overflow, but it could result in a loss of precision for a very large size_t value. Again, it doesn't make a lot of sense to convert a size_t to a double ; you're still better off keeping the value in a size_t variable.

( R Sahu's answer has some suggestions if you can't avoid the cast, such as throwing an exception on overflow.)

Answer 2

If your code is prepared to deal with overflow errors, you can throw an exception if data is too large.

size_t data = 99999999;
if ( data > INT_MAX )
{
   throw std::overflow_error("data is larger than INT_MAX");
}
int convertData = static_cast<int>(data);

Answer 3

静态投射：

static_cast<int>(data);

Answer 4

You can use Boost numeric_cast .

This throws an exception if the source value is out of range of the destination type, but it doesn't detect loss of precision when converting to double .

Whatever function you use, though, you should decide what you want to happen in the case where the value in the size_t is greater than INT_MAX . If you want to detect it use numeric_cast or write your own code to check. If you somehow know that it cannot possibly happen then you could use static_cast to suppress the warning without the cost of a runtime check, but in most cases the cost doesn't matter anyway.

Answer 5

Assuming that the program cannot be redesigned to avoid the cast (ref. Keith Thomson's answer ):

To cast from size_t to int you need to ensure that the size_t does not exceed the maximum value of the int. This can be done using std::numeric_limits :

int SizeTToInt(size_t data)
{
    if (data > std::numeric_limits<int>::max())
        throw std::exception("Invalid cast.");
    return std::static_cast<int>(data);
}

If you need to cast from size_t to double, and you need to ensure that you don't lose precision, I think you can use a narrow cast (ref. Stroustrup: The C++ Programming Language, Fourth Edition):

template<class Target, class Source>
Target NarrowCast(Source v)
{
    auto r = static_cast<Target>(v);
    if (static_cast<Source>(r) != v)
        throw RuntimeError("Narrow cast failed.");
    return r;
}

I tested using the narrow cast for size_t-to-double conversions by inspecting the limits of the maximum integers floating-point-representable integers (code uses googletest):

EXPECT_EQ(static_cast<size_t>(NarrowCast<double>(size_t{ IntegerRepresentableBoundary() - 2 })), size_t{ IntegerRepresentableBoundary() - 2 });
EXPECT_EQ(static_cast<size_t>(NarrowCast<double>(size_t{ IntegerRepresentableBoundary() - 1 })), size_t{ IntegerRepresentableBoundary() - 1 });
EXPECT_EQ(static_cast<size_t>(NarrowCast<double>(size_t{ IntegerRepresentableBoundary() })), size_t{ IntegerRepresentableBoundary() });
EXPECT_THROW(NarrowCast<double>(size_t{ IntegerRepresentableBoundary() + 1 }), std::exception);
EXPECT_EQ(static_cast<size_t>(NarrowCast<double>(size_t{ IntegerRepresentableBoundary() + 2 })), size_t{ IntegerRepresentableBoundary() + 2 });
EXPECT_THROW(NarrowCast<double>(size_t{ IntegerRepresentableBoundary() + 3 }), std::exception);
EXPECT_EQ(static_cast<size_t>(NarrowCast<double>(size_t{ IntegerRepresentableBoundary() + 4 })), size_t{ IntegerRepresentableBoundary() + 4 });
EXPECT_THROW(NarrowCast<double>(size_t{ IntegerRepresentableBoundary() + 5 }), std::exception);

where

constexpr size_t IntegerRepresentableBoundary()
{
    static_assert(std::numeric_limits<double>::radix == 2, "Method only valid for binary floating point format.");
    return size_t{2} << (std::numeric_limits<double>::digits - 1);
}

That is, if N is the number of digits in the mantissa, for doubles smaller than or equal to 2^N, integers can be exactly represented. For doubles between 2^N and 2^(N+1), every other integer can be exactly represented. For doubles between 2^(N+1) and 2^(N+2) every fourth integer can be exactly represented, and so on.